Given $\dfrac{x^3 + 12x}{6x^2 + 8} = \dfrac{y^3 + 27y}{9y^2 + 27}.$ Using componendo and dividendo, find x : y.

Given,

$\dfrac{x^3 + 12x}{6x^2 + 8} = \dfrac{y^3 + 27y}{9y^2 + 27}$

By componendo and dividendo,

$\Rightarrow \dfrac{x^3 + 12x + 6x^2 + 8}{x^3 + 12x - 6x^2 - 8} = \dfrac{y^3 + 27y + 9y^2 + 27}{y^3 + 27y - 9y^2 - 27} \\[1em] \Rightarrow \dfrac{x^3 + (3 \times x \times 2^2 ) + (3 \times x^2 \times 2 ) + 2^3}{x^3 + (3 \times x \times 2^2 ) - (3 \times x^2 \times 2 ) - 2^3} \\[1em] = \dfrac{y^3 + (3 \times y \times 3^2) + (3 \times y^2 \times 3) + 3^3}{y^3 + (3 \times y \times 3^2) - (3 \times y^2 \times 3) - 3^3} \\[1em] \Rightarrow \Big(\dfrac{x + 2}{x - 2}\Big)^3 = \Big(\dfrac{y + 3}{y - 3}\Big)^3 \\[1em] \Rightarrow \dfrac{x + 2}{x - 2} = \dfrac{y + 3}{y - 3}$

Again applying componendo and dividendo,

$\Rightarrow \dfrac{x + \cancel{2} + x - \cancel{2}}{\cancel{x} + 2 - \cancel{x} + 2} = \dfrac{y + \cancel{3} + y - \cancel{3}}{\cancel{y} + 3 - \cancel{y} + 3} \\[1em] \Rightarrow \dfrac{2x}{4} = \dfrac{2y}{6} \\[1em] \Rightarrow \dfrac{x}{y} = \dfrac{2}{6} \times \dfrac{4}{2} \\[1em] \Rightarrow \dfrac{x}{y} = \dfrac{2}{3} \\[1em] \Rightarrow x : y = 2 : 3.$

Hence, the value of ratio x : y is 2 : 3.